CM $2113$ Fall $2024$

A grader costs $$450,000$ to purchase and is expected to have a useful life of $10$ years. Annual

operating and maintenance costs are estimated to be $$30,000$ per year, and the salvage value

after $10$ years of use is estimated to be $$85,000.$ At an interest rate of $6\%$ what is the present

worth equivalent cost to the contractor of owning and operating the grader?

Solution:

Purchase Price: PP $=450,000$

Annual Cost: $A=30,000$

Salvage Value: $SV=85,000$

$N=10$ years

Present Total Cost:

$P_{\text{t}\text{o}\text{t}\text{a}\text{l}}=PP+A^{**}(\frac{P}{A},6\%,10)-S{V}^{**}(\frac{P}{F},6\%,10)$

$=450,000+30,000**7\mathrm{.}360-85,000**0\mathrm{.}558$

$=623,370$

$i=6\%$

A grader costs $$450,000$ to purchase and is expected to have a useful life of $10$ years. Annual

operating and maintenance costs are estimated to be $$30,000$ per year, and the salvage value

after $10$ years of use is estimated to be $$85,000.$ At an interest rate of $6\%$ what is the present

worth equivalent cost to the contractor of owning and operating the grader?

Solution:

Purchase Price: $PP=450,000$

Annual cost: $A=30,000$

Salvage Value: $SV=85,000$

$N=10$ years

$i=6\%$

$=623,370$

Present Total Cost:

$P_{\text{t}\text{o}\text{t}\text{a}\text{l}}=PP+A^{**}(\frac{P}{A},6\%,10)-S{V}^{**}(\frac{P}{F},6\%,10)$

$=450,000+30,000**7\mathrm{.}360-85,000**0\mathrm{.}558$